Proper map

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In mathematics, a function between topological spaces is called proper if inverse images of compact subsets are compact.Template:Sfn In algebraic geometry, the analogous concept is called a proper morphism.

Definition

There are several competing definitions of a "proper function". Some authors call a function $f : X \to Y$ between two topological spaces Template:Em if the preimage of every compact set in $Y$ is compact in $X .$ Other authors call a map $f$ Template:Em if it is continuous and Template:Em; that is if it is a continuous closed map and the preimage of every point in $Y$ is compact. The two definitions are equivalent if $Y$ is locally compact and Hausdorff. Template:Collapse top Let $f : X \to Y$ be a closed map, such that $f^{- 1} (y)$ is compact (in $X$ ) for all $y \in Y .$ Let $K$ be a compact subset of $Y .$ It remains to show that $f^{- 1} (K)$ is compact.

Let ${U_{a} : a \in A}$ be an open cover of $f^{- 1} (K) .$ Then for all $k \in K$ this is also an open cover of $f^{- 1} (k) .$ Since the latter is assumed to be compact, it has a finite subcover. In other words, for every $k \in K,$ there exists a finite subset $γ_{k} \subseteq A$ such that $f^{- 1} (k) \subseteq \cup_{a \in γ_{k}} U_{a} .$ The set $X ∖ \cup_{a \in γ_{k}} U_{a}$ is closed in $X$ and its image under $f$ is closed in $Y$ because $f$ is a closed map. Hence the set $V_{k} = Y ∖ f (X ∖ \cup_{a \in γ_{k}} U_{a})$ is open in $Y .$ It follows that $V_{k}$ contains the point $k .$ Now $K \subseteq \cup_{k \in K} V_{k}$ and because $K$ is assumed to be compact, there are finitely many points $k_{1}, \dots, k_{s}$ such that $K \subseteq \cup_{i = 1}^{s} V_{k_{i}} .$ Furthermore, the set $Γ = \cup_{i = 1}^{s} γ_{k_{i}}$ is a finite union of finite sets, which makes $Γ$ a finite set.

Now it follows that $f^{- 1} (K) \subseteq f^{- 1} (\cup_{i = 1}^{s} V_{k_{i}}) \subseteq \cup_{a \in Γ} U_{a}$ and we have found a finite subcover of $f^{- 1} (K),$ which completes the proof. Template:Collapse bottom

If $X$ is Hausdorff and $Y$ is locally compact Hausdorff then proper is equivalent to Template:Em. A map is universally closed if for any topological space $Z$ the map $f \times {id}_{Z} : X \times Z \to Y \times Z$ is closed. In the case that $Y$ is Hausdorff, this is equivalent to requiring that for any map $Z \to Y$ the pullback $X \times_{Y} Z \to Z$ be closed, as follows from the fact that $X \times_{Y} Z$ is a closed subspace of $X \times Z .$

An equivalent, possibly more intuitive definition when $X$ and $Y$ are metric spaces is as follows: we say an infinite sequence of points ${p_{i}}$ in a topological space $X$ Template:Em if, for every compact set $S \subseteq X$ only finitely many points $p_{i}$ are in $S .$ Then a continuous map $f : X \to Y$ is proper if and only if for every sequence of points ${p_{i}}$ that escapes to infinity in $X,$ the sequence ${f (p_{i})}$ escapes to infinity in $Y .$

Properties

Every continuous map from a compact space to a Hausdorff space is both proper and closed.
Every surjective proper map is a compact covering map.
- A map $f : X \to Y$ is called a Template:Em if for every compact subset $K \subseteq Y$ there exists some compact subset $C \subseteq X$ such that $f (C) = K .$
A topological space is compact if and only if the map from that space to a single point is proper.
If $f : X \to Y$ is a proper continuous map and $Y$ is a compactly generated Hausdorff space (this includes Hausdorff spaces that are either first-countable or locally compact), then $f$ is closed.^[1]